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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Shannon Diversity Index One way to measure species diversity is to use the Shannon diversity index H. A habitat consists of three species A. B. and C, and its Shannon diversity index is

H = x ln x y ln y z ln z where x is the percent of species A in the habitat, y is the percent of species B in the habitat, and z is the percent of species C in the habitat. Use the fact that x + y   +   z =   1

(the sum of the three must equal 100 %

) to show that the maximum value of H occurs when

x = y = z = 1 3

What is the maximum value of H?

To determine

To calculate: The value of percent of different species of habitant as x=y=z=13 for maximum for maximum value of Shannon diversity index H.

Explanation

Given Information:

Shannon diversity index is,

H=xlnxylnyzlnz

Where H is Shannon diversity index,

x is percent of species A in habitant,

y is percent of species B in habitant.

z is percent of species C in habitant.

and also given,

x+y+z=1

We have to show that H is maximum at x=y=z=13.

Formula used:

H=xlnxylnyzlnz

H is Shannon diversity index.

x+y+z=1z=1xy

Substitute z=1xy in H.

H=xlnxylny(1xy)ln(1xy)

Is the primary equation.

How to maximize H, following procedure should be followed:

Step 1: Write the primary equation in term of two variable only, to maximize it.

Step 2: Find first partial derivative of primary equation and equate to zero, to find critical

point like:

Hx(x,y)=0Hy(x,y)=0

Step 3: Find double partial derivative with respect to all variables separately at different

critical points.

For critical point (x,y)(a,b).

Double partial derivative can be written as,

Hxx(a,b)y:Hyy(a,b)Hxy(a,b)andHyx(x,y)

Then to text the function gives maximum, minimum and saddle point value,

Consider the secondary equation:

d=Hxx(a,b)Hyy(a,b)[Hxy(a,b)]2

If d>0 and Hxx(a,b)>0 then function.

H has a minimum value at (a,b),

If d>0 and Hxx<0 then function,

H has a maximum value at (a,b),

If d<0, then [(a,b),H(x,y)] is a saddle point.

The test gives no information when d=0.

Calculation:

Consider the primary equation,

H(x,y)=xlnxylny(1xy)ln(1xy).

Partially differentiate H(x,y) with respect to x, keeping y as constant.

Hx(x,y)=n[xlnxylny(1xy)ln(1xy)]Hx(x,y)=lny1(1)ln(1xy)(1xy)(1xy)(1)Hy(x,y)=ln(1xyy)

Now equating Hx(x,y)andHy(x,y) to zero.

For critical value (x,y)(a,b)

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